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Intrinsic Linking and Knotting Are Arbitrarily Complex in Directed Graphs

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Intrinsic Linking and Knotting Are Arbitrarily Complex in Directed Graphs INTRINSIC LINKING AND KNOTTING ARE ARBITRARILY COMPLEX IN DIRECTED GRAPHS THOMAS W. MATTMAN, RAMIN NAIMI, AND BENJAMIN PAGANO Abstract. Fleming and Foisy [4] recently proved the existence of a digraph whose every embedding contains a 4-component link, and left open the possibility that a directed graph with an intrin- sic n-component link might exist. We show that, indeed, this is the case. In fact, much as Flapan, Mellor, and Naimi [2] show for graphs, knotting and linking are arbitrarily complex in directed graphs. Specifically, we prove the analog for digraphs of the main theorem of their paper: for any n and α, every embedding of a suffi- ciently large complete digraph in R3 contains an oriented link with components Q1;:::;Qn such that, for every i 6= j, jlk(Qi;Qj)j ≥ α and ja2(Qi)j ≥ α, where a2(Qi) denotes the second coefficient of the Conway polynomial of Qi. 1. Introduction Fleming and Foisy [4] recently proved the existence of a digraph whose every embedding contains a 4-component link, and left open the possibility that a directed graph with an intrinsic n-component link might exist. We show that, indeed, this is the case. In fact, much as Flapan, Mellor, and Naimi [2] show for graphs, knotting and linking are arbitrarily complex in directed graphs. Specifically, we prove the analog for digraphs of the two main theorems of their paper. Before stating the results, we introduce some notation. For graph G, the symmetric digraph DG is obtained by replacing each edge vivj with two directed edges, vivj and vjvi. For cycle C in a digraph, let p1; : : : ; pδ arXiv:1901.01212v1 [math.GT] 4 Jan 2019 in C be maximal consistently directed paths (no pi is a subpath of a longer consistently directed path in C). Then δ is the directionality of C. Thus, a consistently directed cycle is 1{directional. Following [2], we define the linking pattern of a link of n components, L1;:::;Ln, as the weighted graph on vertices v1; : : : ; vn where jlk(Li;Lj)j (if nonzero) Date: January 7, 2019. 2010 Mathematics Subject Classification. Primary 05C10, Secondary 57M15, 57M25, 05C20, 05C35 . Key words and phrases. intrinsically knotted graph, intrinsically linked graph, directed graph, spatial graph. 1 2 THOMAS W. MATTMAN, RAMIN NAIMI, AND BENJAMIN PAGANO is the weight of vivj. When the linking number is zero, there is no edge. The mod 2 linking pattern instead carries the weights !(Li;Lj) = lk(L1;Lj) mod 2. Theorem 1. Let λ, δ 2 N with δ even or 1. For every n 2 N, there is a digraph DG such that every embedding of DG in R3 contains a link whose weighted linking pattern is Kn with every weight at least λ and every component δ-directional. Theorem 2. For every n; α 2 N, there is a complete digraph DKr such 3 that every embedding of DKr in R contains a link with 1-directional components Q1;:::;Qn such that for every i 6= j, jlk(Qi;Qj)j ≥ α and ja2(Qi)j ≥ α. Not just the statements of our theorem are similar to [2], but the proofs as well. We prove Theorem 1 in the next section and Theorem 2 in Section 3. 2. Intrinsic Linking In this section we prove Theorem 1. After a couple of introductory lemmas, we follow the same path as in [2]. Throughout this paper, indices are cyclic; e.g., given say xi with 1 ≤ i ≤ n (or 0 ≤ i ≤ n), xn+1 is to be understood as x1 (x0); and, more generally, for i > n, xi is to be understood as xi−n (xi−n−1). Lemma 1. Every spatial digraph DK6m contains at least m pairwise disjoint 2-component links, all with odd linking numbers, such that all their components are 2-directional. Proof. Take an undirected graph K6 with vertices v1; : : : ; v6 and form the digraph DG by orienting the edges such that vivj is directed from vi to vj when i < j. Note that DG contains no 1-directional cycles. Additionally, notice that all 3-cycles in DG must be 2-directional. As argued in [1, 3], any embedding of K6 contains a pair of 3-cycles with odd linking number. Then, any embedding of DG must contain a link with 2-directional components. As DG is a subgraph of DK6 and DK6m contains m distinct copies of DK6, every embedding of DK6m in R3 contains m pairwise disjoint links with odd linking number and 2-directional components. Lemma 2. Let M be an m × n matrix with entries in Z2 where every column of M contains at least one 1. For every n and m, there exists n a vector v 2 row(M) for which over 2 of the entries are 1's. LINKING AND KNOTTING IN DIGRAPHS 3 Proof. Let kvk denote the number of 1's in a vector v. We proceed by induction on n. If n = 1, the statement is obvious. Fix m > 0 and assume that for all 1 ≤ l < n, the statement holds for every m × l matrix. Let m be an m × n matrix and suppose v0 is a vector in the row space that maximizes kv0k. For a contradiction, n assume that kv0k = k ≤ 2 . Without loss of generality, the first k entries in v0 are 1, and the rest are all 0. Divide M into the two matrices ML and MR where the m×k matrix ML is the first k columns of M and the m × n − k MR is the remaining n − k columns of M. Similarly, split vectors x 2 row(M) into vectors xL and xR of lengths k and n − k respectively. For every x, kxLk ≥ kxRk, because otherwise kv0 + xk > kv0k, contradicting the maximality of kv0k. By induction, n−k n−k there is an x such that kxRk > 2 . This would imply kxLk > 2 , and kxk > n − k ≥ k = kv0j, a contradiction. Lemma 3. Suppose a spatial digraph DKp contains links with com- ponents J1; ··· ;J2n and X1; ··· X2n such that Ji is 2-directional and !(Ji;Xi) = 1 for every i ≤ n. Then DKp contains a 1-directional cycle Z in DKp with vertices on J1 [···[ J2n such that for some I ⊆ f1; ··· ; 2ng with jIj ≥ n=2, !(Z; Xi) = 1 for all i 2 I. Further- more, for every δ ≥ 1, Z can be chosen to be 2δ-directional if DKp contains at least 2δ − 2 vertices disjoint from all Ji and Xi. Z J J1 J2 n u w u1 w1 u2 w2 n n X1 X2 Xn Xi Xi Xi 1 2 n 2 Figure 1. Illustration of Lemma 3 Proof. Since each Ji is 2-directional, it has exactly two vertices where \direction changes" on Ji, i.e., the two edges at each of these vertices are directed either both toward or both away from that vertex. We label these two vertices ui and wi, so that both paths on Ji between ui and wi are directed from ui toward wi. Let qi be one of these two 4 THOMAS W. MATTMAN, RAMIN NAIMI, AND BENJAMIN PAGANO directed paths; let wiui+1 denote the directed edge from wi to ui+1; and S S let C = qi [ wiui+1. Note that C is a 1-directional cycle. If we want the cycle Z in the conclusion of the lemma to be 2 directional, we make C 2-directional by replacing the edge w2nu1 in C with the edge u1w2n. And if we want Z to be 2δ directional with δ ≥ 2, we replace the edge w2nu1 in C with a path between w2n and u1 that goes through the 2δ − 2 extra vertices given in the hypothesis of the lemma, such that the path changes direction at each of those vertices and also at w2n and u1. If !(C; Xi) = 1 for at least n=2 of the Xj's, we let Z = C, and we are done. Otherwise, we construct Z as follows. Let M be the matrix with entries Mij = !(Ji;Xj). Since Mii = 1 for all i, by Lemma 2, there exist rows i1; ··· ; ik of M whose sum contains greater than n 1's. Let Z = CrJi1 · · · rJik . Since C links fewer than n=2 of the Xj's, while the sum of the rows i1; ··· ; ik contains greater than n 1's, it follows that Z links at least n=2 of the Xj's, as desired. Recall from [2] that a generalized mod 2 keyring link is one whose mod 2 linking pattern includes an n-star. Proposition 1. Let n; δ; and 2 N with each of δ and either even or 1. There is a digraph DG such that every embedding of DG in R3 contains a link whose mod 2 linking pattern contains the complete bipartite graph Kn;n, where every component of the first partition is δ- directional and every component of the second partition is -directional. Proof. The argument largely follows the proof of the corresponding proposition in [2], and we begin by summarizing their approach. First observe that for any given m, there's a p such that every embedding of Kp contains m disjoint mod 2 generalized keyrings, each having n keys. Let X1;:::;Xm denote the rings. Apply (the analogue of) Lemma 3 n times to construct n cycles Z1;:::;Zn and index set In of size at least n so that, for every i 2 In and every j ≤ n, !(Zj;Xi) = 1.
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